Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 29

Answer

$$\eqalign{ & {\bf{T}}\left( t \right) = - \sin \omega t{\bf{i}} + \cos \omega t{\bf{j}} \cr & {\bf{N}}\left( t \right) = - \cos \omega t{\bf{i}} - \sin \omega t{\bf{j}} \cr & {a_{\bf{T}}} = 0 \cr & {a_{\bf{N}}} = a{\omega ^2} \cr} $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = a\cos \omega t{\bf{i}} + a\sin \omega t{\bf{j}} \cr & {\text{Calculate }}{\bf{v}}\left( t \right){\text{ and }}{\bf{a}}\left( t \right) \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {a\cos \omega t{\bf{i}} + a\sin \omega t{\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = - a\omega \sin \omega t{\bf{i}} + a\omega \cos \omega t{\bf{j}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ { - a\omega \sin \omega t{\bf{i}} + a\omega \cos \omega t{\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = - a{\omega ^2}\cos \omega t{\bf{i}} - a{\omega ^2}\sin \omega t{\bf{j}} \cr & {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr & {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left\| {{\bf{v}}\left( t \right)} \right\|}},{\text{ }}{\bf{v}}\left( t \right) \ne 0 \cr & {\bf{T}}\left( t \right) = \frac{{ - a\omega \sin \omega t{\bf{i}} + a\omega \cos \omega t{\bf{j}}}}{{\sqrt {{{\left( { - a\omega \sin \omega t} \right)}^2} + {{\left( {a\omega \cos \omega t} \right)}^2}} }} \cr & {\bf{T}}\left( t \right) = \frac{{ - a\omega \sin \omega t{\bf{i}} + a\omega \cos \omega t{\bf{j}}}}{{\sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t} }} \cr & {\bf{T}}\left( t \right) = \frac{{ - a\omega \sin \omega t{\bf{i}} + a\omega \cos \omega t{\bf{j}}}}{{\sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right)} }} \cr & {\bf{T}}\left( t \right) = \frac{{ - a\omega \sin \omega t{\bf{i}} + a\omega \cos \omega t{\bf{j}}}}{{a\omega }} \cr & {\bf{T}}\left( t \right) = - \sin \omega t{\bf{i}} + \cos \omega t{\bf{j}} \cr & {\text{Find }}{\bf{T}}'\left( t \right) \cr & {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ { - \sin \omega t{\bf{i}} + \cos \omega t{\bf{j}}} \right] \cr & {\bf{T}}'\left( t \right) = - \omega \cos \omega t{\bf{i}} - \omega \sin \omega t{\bf{j}} \cr & \left\| {{\bf{T}}'\left( t \right)} \right\| = \sqrt {{{\left( { - \omega \cos \omega t} \right)}^2} + {{\left( { - \omega \sin \omega t} \right)}^2}} \cr & \left\| {{\bf{T}}'\left( t \right)} \right\| = \sqrt {{\omega ^2}{{\cos }^2}\omega t + {\omega ^2}{{\sin }^2}\omega t} \cr & \left\| {{\bf{T}}'\left( t \right)} \right\| = \omega \cr & {\text{Finding the Principal Unit Normal Vector}} \cr & {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr & {\bf{N}}\left( t \right) = \frac{{ - \omega \cos \omega t{\bf{i}} - \omega \sin \omega t{\bf{j}}}}{\omega } \cr & {\bf{N}}\left( t \right) = - \cos \omega t{\bf{i}} - \sin \omega t{\bf{j}} \cr & {\text{Find }}{a_{\bf{T}}}{\text{ and }}{a_{\bf{N}}} \cr & {a_{\bf{T}}} = {\bf{a}} \cdot {\bf{T}} = \left( { - a{\omega ^2}\cos \omega t{\bf{i}} - a{\omega ^2}\sin \omega t{\bf{j}}} \right)\left( { - \sin \omega t{\bf{i}} + \cos \omega t{\bf{j}}} \right) \cr & {a_{\bf{T}}} = {\bf{a}} \cdot {\bf{T}} = a{\omega ^2}\sin \omega t\cos \omega t - a{\omega ^2}\sin \omega t\cos \omega t \cr & {a_{\bf{T}}} = 0 \cr & {a_{\bf{N}}} = {\bf{a}} \cdot {\bf{N}} = \left( { - a{\omega ^2}\cos \omega t{\bf{i}} - a{\omega ^2}\sin \omega t{\bf{j}}} \right)\left( { - \cos \omega t{\bf{i}} - \sin \omega t{\bf{j}}} \right) \cr & {a_{\bf{N}}} = a{\omega ^2}{\cos ^2}\omega t + a{\omega ^2}{\sin ^2}\omega t \cr & {a_{\bf{N}}} = a{\omega ^2} \cr} $$
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